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I'm reading Stillwell's Mathematics and Its History. In the chapter about pythagorean triples, he says that we now know that the general formula for generating pythagorean triples is:

$$a = r\cdot(m^2 - n^2) ,\ \, b = r\cdot(2mn) ,\ \, c = r\cdot(m^2 + n^2)\tag{1}$$

And that there is a special case which gives all solutions $a,b,c$ without common divisor:

$$a = m^2 - n^2 ,\ \, b = 2mn ,\ \, c = m^2 + n^2\tag{2}$$

I'm just no so sure where are these formulas coming from. I've made the following trial: By looking $(2)$, I've written the formula for the pythagorean theorem and I've switched only the $c$:

$$a^2+b^2=\color{red}{(m^2 + n^2)^2}$$

And expanded it:

$$a^2+b^2=\color{red}{m^4+2n^2m^2+n^4}\tag{3}$$

Then my guess is that I should find arbitrary $a$, $b$ (and perhaps some integers?) written in terms of $n$ and $m$ such that $(3)$ is true. So I guess I need to split the polynomial in (perhaps) some arbitrary way such as:

$$a^2+b^2=\color{red}{m^4+2n^2m^2}\color{green}{+n^4}$$ $$a^2+b^2=\color{red}{m^4+n^2m^2}\color{green}{+n^2m^2+n^4}$$

And then find $a,b$ such that:

$$\begin{eqnarray*} {a^2}&=&{m^4+2n^2m^2} \\ {b^2}&=&{n^4} \end{eqnarray*}$$

Or:

$$\begin{eqnarray*} {a^2}&=&{m^4+n^2m^2} \\ {b^2}&=&{n^2m^2+n^4} \end{eqnarray*}$$

Or any arbitrary combination such that $(3)$ holds. I'm still a little stuck, am I on the right path? I've looked wikipedia article on pythagorean triples and there's no much insight into this matter.

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From where ? From Euclid. :-) –  Lucian Mar 16 at 23:44
    
@Lucian Yes. The worst part is that it's all in there. I've seen "proof" and somehow I thought that it wouldn't answer my question. Where's the "I feel ashamed for asking such a dumb question and want to delete it to avoid future embarassment" button? –  Vladimir Putin Mar 16 at 23:56
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This is possibly a duplicate of this question. –  robjohn Mar 16 at 23:57
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@Pristine, don't worry about it. We all sometimes feel like we're asking dumb questions. But really, the only dumb questions are the ones you don't ask. –  David Mar 18 at 5:28

3 Answers 3

up vote 3 down vote accepted

If you read a few pages further you will find that Exercises 1.31. and 1.3.2 give a derivation of the Pythagorean triple formula using a rational parametrization of the circle (see also the "geometry" section of the Wikipedia article).

More generally, if a conic curve with rational coefficients has one rational point $\rm\:P\:$ then it has infinitely many, since any rational line through $\rm\:P\:$ will intersect the curve in another point, necessarily rational, since if one root of a rational quadratic is rational then so is the other. Therefore, by sweeping lines of varying rational slopes through $\rm\:P\:$ we obtain infinitely many rational points on the conic. Projecting these points onto a line leads to a rational parametrization of the conic. For a very nice exposition see Chapter $1$ of Silverman and Tate: Rational Points on Elliptic Curves. There are also many online expositions, e.g. search on "rational parametrization Pythagorean triple".

There are also many other ways to derive the formula. A particularly beautiful one-line way employs Hilbert's Theorem $90$ - see the award winning expository Monthly paper by Olga Taussky Sums of Squares.

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I got sad for asking such a stupid question - after you said the answer is in wikipedia, I noticed it really was. But when you answered the question and provided references, I got really happy! I'm a books colector, then answers with references are a supreme delight, specially coming from a very qualified member of the mathematical community like you. Thanks! If you have some blog or website where you recommend books and articles, please let me know. –  Vladimir Putin Mar 17 at 2:48
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Don't worry, we all pondered this question at one point on our learning curve. The geometrical approach is very pretty so I recommend that you read about it. Another beautiful geometrical viewpoint is the reflective generation of the ternary tree of primitive Pythagorean triples due to Aubry. –  Bill Dubuque Mar 17 at 3:00

There is a fair bit of work to do to prove this theorem but you could look here, starting on page 2. The article is aimed at a secondary school level audience, so hopefully it should be fairly accessible.

Note that the main difficulty is not proving that the formulae give some possible $a,b,c$ but proving that they give all possible $a,b,c$.

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It comes from a rather sneaky relation that $2m^2n^2=4m^2n^2-2m^2n^2$. If we have $c=m^2+n^2$, then $$c^2=(m^2+n^2)^2=m^4+2m^2n^2+n^4=\\m^4-2m^2n^2+n^4+4m^2n^2=(n^2-m^2)^2+(2mn)^2=a^2+b^2$$ where $a=(m^2+n^2),b=2mn$.

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That doesn't actually prove anything since you're assuming a certain form for $c$. –  nbubis Mar 17 at 0:19

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